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Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Solution:

Let the side of the first square be x and the side of the second square be y.

Area of a square = side × side

Perimeter of a square = 4 × side

Therefore, the area of the first and second square is x2 and y2 respectively.

Also, the perimeters of the first and second square are 4x and 4y respectively.

Applying the known conditions:

(i) x2 + y2 = 468 ...(1)

(ii) 4x - 4y = 24 ....(2) considering x > y

4(x - y) = 24

x - y = 6

x = 6 + y

Substitute x = y + 6 in equation (1)

(y + 6)2 + y2 = 468

y2 +12y + 36 + y2 = 468

2y2 + 12y + 36 = 468

2(y2 + 6y + 18) = 468

y2 + 6y + 18 = 234

y2 + 6y + 18 - 234 = 0

y2 + 6y - 216 = 0

Solving by factorization method,

y2 + 18y - 12y - 216 = 0

y (y + 18) - 12 (y + 18) = 0

(y + 18) (y - 12) = 0

y + 18 = 0 and y - 12 = 0

y = -18 and y = 12

y cannot be a negative value as it represents the side of the square.

Side of the first square x = y + 6 = 12 + 6 = 18 m

Side of the second square = 12 m.

☛ Check: Class 10 Maths NCERT Solutions Chapter 4

Video Solution:

Sum of the areas of two squares 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Class 10 Maths NCERT Solutions Chapter 4 Exercise 4.3 Question 11

Summary:

Sum of the areas of two squares 468 m², if the difference of perimeters is 24 m, then the side of the first square is 18 m and the side of the second square is 12 m.

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